Question

One end of string of length $l$ is connected to a particle of mass $m$ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed $v$, the net force on the particle (directed towards centre) will be ($T$ represents the tension in the spring)
(A) $zero$
(B) $T$
(C) \[T + \dfrac{{m{v^2}}}{l}\]
(D) \[T - \dfrac{{m{v^2}}}{l}\]

Answer 1

Hint:Here, a particle of mass $m$ is performing a circular motion and is connected to string. The string is attached to a peg so that the particle is confined in the circle of radius \[l\]. Now, you are asked to find the net force acting on the particle in the direction towards the centre. You know that an object performs a circular motion when a centripetal force acts on the body, you need to find that centripetal force and also its source.

Complete answer:
Picturise the situation given to you. The string is attached to the peg on one end and the other end is attached to the particle. The particle is performing circular motion. Obviously, a centripetal force must be acting on the particle and that is the reason the particle is in circular motion. Now, if you think, the only source of force that you can see is the force due to the string on the particle, that is the tension force. So, the centripetal force equal to $\dfrac{{m{v^2}}}{l}$ is acting on the particle which is equal to the tension $ \to \dfrac{{m{v^2}}}{l} = T$.
Therefore, the net force on the particle (directed towards centre) will be $T$.

Hence, option B is correct.

Note: Always remember that an object performs a circular motion only if the object is in the influence of a centripetal force. In the above question, the source of the centripetal force was the tension force acting on the particle. In case of earth revolving around the sun, it is the gravitational force that is acting on earth due to the sun. Also keep in mind that the magnitude of centripetal force is equal to $\dfrac{{m{v^2}}}{l}$ and its direction is always towards the centre.